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(C) In the steady state,the energy absorbed by the middle plate is equal to the energy released by the middle plate.Let $T'$ be the temperature of the

Vedclass · Accepted Answer

${\left( {\frac{{97}}{2}} \right)^{\frac{1}{4}}}\,T$

Vedclass · Answer

(C) In the steady state,the energy absorbed by the middle plate is equal to the energy released by the middle plate.Let $T'$ be the temperature of the middle plate.The heat flux absorbed from the hotter plate (at $3T$) is $\sigma A(3T)^4 - \sigma A(T')^4$.The heat flux released to the colder plate (at $2T$) is $\sigma A(T')^4 - \sigma A(2T)^4$.Equating the two:$\sigma A(3T)^4 - \sigma A(T')^4 = \sigma A(T')^4 - \sigma A(2T)^4$$(3T)^4 - (T')^4 = (T')^4 - (2T)^4$$81T^4 + 16T^4 = 2(T')^4$$97T^4 = 2(T')^4$$(T')^4 = \frac{97}{2}T^4$$T' = \left( \frac{97}{2} \right)^{\frac{1}{4}} T$

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